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Determining when a function is an element of L^2 (0,1)

  • Thread starter lants
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  • #1
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Homework Statement


Determine for what k f(x)=xk is an element of L2 (0,1) vector space
k ∈ ℝ

Homework Equations




The Attempt at a Solution


[tex] \int_{0}^{1} x^{2k} dx = \frac{1-0^{2k+1}}{1+2k} = \sum_{n=0}^{\infty}{(-2k)^{n}} [/tex] (for k > -½)

This sum should converge for [tex]
\lim_{n \to +\infty}
{\frac{|(-2k)^{n+1}|}{|(-2k)^{n}|}} < 1
=
|-2k| < 1
[/tex]

Which gives me a radius of convergence for
[tex] - \frac{1}{2} < k < \frac{1}{2}
[/tex]

But just by examining it, the integral should exist for any k greater than negative one-half, what is wrong with my ratio test?
 
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Answers and Replies

  • #2
LCKurtz
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Homework Statement


Determine for what k f(x)=xk is an element of L2 (0,1) vector space
k ∈ ℝ

Homework Equations




The Attempt at a Solution


[tex] \int_{0}^{1} x^{2k} dx = \frac{1-0^{2k+1}}{1+2k} = \sum_{n=0}^{\infty}{(-2k)^{n}} [/tex] (for v > -½)

This sum should converge for [tex]
\lim_{n \to +\infty}
{\frac{|(-2k)^{n+1}|}{|(-2k)^{n}|}} < 1
=
|-2k| < 1
[/tex]

Which gives me a radius of convergence for
[tex] - \frac{1}{2} < k < \frac{1}{2}
[/tex]

But just by examining it, the integral should exist for any k greater than negative one-half, what is wrong with my ratio test?
The infinite series doesn't represent the fraction for all values of k. The series may not converge for some k but that doesn't mean the fraction's value doesn't exist.
 
  • #3
14
1
Ok thanks, obvious mistake
 
  • #4
HallsofIvy
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I don't see your logic. The integral of [itex]x^{2k}[/itex] is the number 1/(2k+ 1) as long as that number exists. What does that have to do with whether or not there is a geometric sequence converging to it? If k= 1, which is greater than 1/2, f(x)= x which is certainly twice integrable between 0 and 1.
 
  • #5
14
1
someone already responded pointing out it was wrong in an actually helpful way, and I already responded back, so move along now
 
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